1 00:00:00,000 --> 00:00:00,940 2 00:00:00,940 --> 00:00:04,750 The goal in this video is to essentially prove a pretty 3 00:00:04,750 --> 00:00:05,920 simple result. 4 00:00:05,920 --> 00:00:08,710 And that's that the ratio between the volumes-- let me 5 00:00:08,710 --> 00:00:11,850 write this down-- that the ratio between the volume at 6 00:00:11,850 --> 00:00:17,580 state B and the volume at state A-- so the ratio of that 7 00:00:17,580 --> 00:00:21,930 volume to that volume-- is equal to, in our Carnot cycle, 8 00:00:21,930 --> 00:00:25,620 is equal to the ratio between the volume at state C. 9 00:00:25,620 --> 00:00:28,860 So this volume and that volume. 10 00:00:28,860 --> 00:00:33,920 So volume at C to the volume at D. 11 00:00:33,920 --> 00:00:36,930 So this is what I'm about to embark to prove. 12 00:00:36,930 --> 00:00:39,720 A fairly simple result, that maybe is even, if you look at 13 00:00:39,720 --> 00:00:42,090 this, that looks about right. 14 00:00:42,090 --> 00:00:43,780 So if you're happy just knowing that, you don't have 15 00:00:43,780 --> 00:00:44,850 to watch the rest of the video. 16 00:00:44,850 --> 00:00:47,440 But if you are curious how we get there, I encourage you to 17 00:00:47,440 --> 00:00:49,400 watch it, although it gets a little bit math-y. 18 00:00:49,400 --> 00:00:51,460 But I think the fun part about it, even, it will 19 00:00:51,460 --> 00:00:52,420 one, satisfy you. 20 00:00:52,420 --> 00:00:53,090 That this is true. 21 00:00:53,090 --> 00:00:55,150 But the other thing is, we'll be able to study adiabatic 22 00:00:55,150 --> 00:00:56,890 processes a little bit more. 23 00:00:56,890 --> 00:00:59,820 So just kind of launching off of that, the whole proof 24 00:00:59,820 --> 00:01:03,250 revolves around to this step right here and this step right 25 00:01:03,250 --> 00:01:05,850 here. when we go from D to A. 26 00:01:05,850 --> 00:01:10,090 So by definition, an adiabatic process is one where there's 27 00:01:10,090 --> 00:01:12,060 no transfer of heat. 28 00:01:12,060 --> 00:01:16,530 So our heat transfer in an adiabatic process is 0. 29 00:01:16,530 --> 00:01:19,010 So if we go back to our original definitions-- let me 30 00:01:19,010 --> 00:01:20,650 show you that here. 31 00:01:20,650 --> 00:01:23,150 Right here, at the step and this step, we have 32 00:01:23,150 --> 00:01:25,470 no transfer of heat. 33 00:01:25,470 --> 00:01:26,565 So if we go back to there. 34 00:01:26,565 --> 00:01:28,680 Adiabatic-- we're completely isolated from 35 00:01:28,680 --> 00:01:29,550 the rest of the world. 36 00:01:29,550 --> 00:01:32,220 So there's nothing to transfer heat to or from. 37 00:01:32,220 --> 00:01:36,660 So if we go to our definition, almost, or our first law of 38 00:01:36,660 --> 00:01:42,540 thermodynamics, we know that the change in internal energy 39 00:01:42,540 --> 00:01:48,500 is equal to the heat applied to the system minus the work 40 00:01:48,500 --> 00:01:49,660 done by the system. 41 00:01:49,660 --> 00:01:51,530 And the work done by the system is equal to the 42 00:01:51,530 --> 00:01:54,420 pressure of the system times some change in volume. 43 00:01:54,420 --> 00:01:56,720 At least, maybe it's a very small change in volume, while 44 00:01:56,720 --> 00:01:57,790 the pressure is constant. 45 00:01:57,790 --> 00:02:00,580 But if we're doing a quasi-static process, we can 46 00:02:00,580 --> 00:02:01,670 write this. 47 00:02:01,670 --> 00:02:04,710 Pressure, you can view it as kind of constant for that very 48 00:02:04,710 --> 00:02:06,920 small change in volume. 49 00:02:06,920 --> 00:02:08,300 So that's what we have there. 50 00:02:08,300 --> 00:02:11,165 Now, if it's adiabatic, we know that this is 0. 51 00:02:11,165 --> 00:02:14,960 52 00:02:14,960 --> 00:02:18,650 And if that's 0, we can add P delta V to both sides of this 53 00:02:18,650 --> 00:02:22,645 equation, and we will get that-- this is only true if it 54 00:02:22,645 --> 00:02:27,000 were adiabatic-- that delta U, our change in internal energy, 55 00:02:27,000 --> 00:02:30,530 plus our pressure times our change in 56 00:02:30,530 --> 00:02:34,095 volume, is equal to 0. 57 00:02:34,095 --> 00:02:36,530 And let's see if we can do this somehow, we can do 58 00:02:36,530 --> 00:02:39,020 something with this equation to get to that result that I'm 59 00:02:39,020 --> 00:02:40,450 trying to get to. 60 00:02:40,450 --> 00:02:45,100 So a few videos ago, I proved to you that U, the internal 61 00:02:45,100 --> 00:02:48,050 energy at any point in time-- let me write it here. 62 00:02:48,050 --> 00:02:54,890 The internal energy at any point in time is equal to 3/2 63 00:02:54,890 --> 00:02:58,400 times n times R times T. 64 00:02:58,400 --> 00:03:01,440 Which is also equal to 3/2 times PV. 65 00:03:01,440 --> 00:03:04,520 Now, if I have a change in internal energy, what can 66 00:03:04,520 --> 00:03:05,460 change on this side? 67 00:03:05,460 --> 00:03:07,000 Something must have changed. 68 00:03:07,000 --> 00:03:08,980 Well, 3/2 can't change. 69 00:03:08,980 --> 00:03:09,850 n can't change. 70 00:03:09,850 --> 00:03:12,190 We're not going to change the number of molecules we have. 71 00:03:12,190 --> 00:03:13,990 The universal gas constant can't change. 72 00:03:13,990 --> 00:03:16,050 So the temperature must change. 73 00:03:16,050 --> 00:03:16,930 So there you have it. 74 00:03:16,930 --> 00:03:21,636 You have delta U could be rewritten as delta-- let me do 75 00:03:21,636 --> 00:03:26,930 it in a different color-- delta U could be written as 76 00:03:26,930 --> 00:03:32,810 3/2 n times R times our change in temperature. 77 00:03:32,810 --> 00:03:34,880 And that's why I keep saying in this-- especially when 78 00:03:34,880 --> 00:03:36,770 we're dealing with the situation where all of the 79 00:03:36,770 --> 00:03:39,440 internal energy is essentially kinetic energy-- that if you 80 00:03:39,440 --> 00:03:40,925 don't have a change in temperature, you're not going 81 00:03:40,925 --> 00:03:42,620 to have a change in internal energy. 82 00:03:42,620 --> 00:03:44,520 Likewise, if you don't have a change in internal energy, 83 00:03:44,520 --> 00:03:47,030 you're not going to have a change in temperature. 84 00:03:47,030 --> 00:03:48,620 So let me put this aside right here. 85 00:03:48,620 --> 00:03:50,850 I'm going to substitute it back there. 86 00:03:50,850 --> 00:03:56,000 But let's see if we can do something with this P here. 87 00:03:56,000 --> 00:04:00,920 Well, we'll just resort to our ideal gas equation. 88 00:04:00,920 --> 00:04:03,470 Because we're dealing with an ideal gas, we might as well. 89 00:04:03,470 --> 00:04:06,720 PV is equal to nRT. 90 00:04:06,720 --> 00:04:10,970 This should be emblazoned in your mind, at this point. 91 00:04:10,970 --> 00:04:15,150 So if we want to solve for P, we get O is 92 00:04:15,150 --> 00:04:19,700 equal to nRT over V. 93 00:04:19,700 --> 00:04:20,250 Fair enough. 94 00:04:20,250 --> 00:04:23,840 So let's put both of these things aside, and substitute 95 00:04:23,840 --> 00:04:25,730 them into this formula. 96 00:04:25,730 --> 00:04:27,580 So delta U is equal to this thing. 97 00:04:27,580 --> 00:04:38,910 So that means that 3/2 nR delta T plus P-- P is this 98 00:04:38,910 --> 00:04:50,740 thing-- plus nRT over V times delta V is equal to 0. 99 00:04:50,740 --> 00:04:51,810 Interesting. 100 00:04:51,810 --> 00:04:53,670 So what can we do further here? 101 00:04:53,670 --> 00:04:56,080 And I'll kind of tell you where I'm going with this. 102 00:04:56,080 --> 00:04:59,570 So that tells me, my change in internal energy over a very 103 00:04:59,570 --> 00:05:03,660 small delta T-- this tells me my work done by the system 104 00:05:03,660 --> 00:05:05,330 over a very small delta V. 105 00:05:05,330 --> 00:05:07,380 And we're saying that, you know, over each little small 106 00:05:07,380 --> 00:05:11,160 increments, they're going to add up to zero. 107 00:05:11,160 --> 00:05:13,740 So let me just go back to the original graph. 108 00:05:13,740 --> 00:05:17,090 So this is over a very small delta V right there. 109 00:05:17,090 --> 00:05:19,060 Let me do it in a more vibrant color. 110 00:05:19,060 --> 00:05:21,560 A very small change, as we go from there 111 00:05:21,560 --> 00:05:22,730 to, let's say, there. 112 00:05:22,730 --> 00:05:24,700 We're going to have some change in our volume. 113 00:05:24,700 --> 00:05:26,590 And you don't see the temperature here. 114 00:05:26,590 --> 00:05:29,030 So don't try to even imagine, when we do the integral, that 115 00:05:29,030 --> 00:05:30,970 we should think of it in some terms of area. 116 00:05:30,970 --> 00:05:32,410 But we're going to integrate over the change in 117 00:05:32,410 --> 00:05:33,820 temperature, as well. 118 00:05:33,820 --> 00:05:35,180 The temperature changes a little bit 119 00:05:35,180 --> 00:05:36,400 from there to there. 120 00:05:36,400 --> 00:05:39,540 So what I want to do-- this is, right here, 121 00:05:39,540 --> 00:05:40,360 over a small change. 122 00:05:40,360 --> 00:05:43,250 I want to integrate eventually over all of the changes that 123 00:05:43,250 --> 00:05:46,470 occur during our adiabatic process. 124 00:05:46,470 --> 00:05:48,650 So let's see if we can simplify this before I break 125 00:05:48,650 --> 00:05:50,110 out the calculus. 126 00:05:50,110 --> 00:05:54,970 So if we divide both sides by nRT, what do we get? 127 00:05:54,970 --> 00:06:01,130 So let's divide it by nRT, let's divide it by nRT. 128 00:06:01,130 --> 00:06:04,770 And we have to do it to both sides of the equation. nRT. 129 00:06:04,770 --> 00:06:08,850 Well, on this term, the n's cancel out, the R cancels out. 130 00:06:08,850 --> 00:06:12,110 Over here, this nRT cancels out with this nRT. 131 00:06:12,110 --> 00:06:13,950 And what are we left with? 132 00:06:13,950 --> 00:06:19,820 We're left with 3/2-- we have this 1 over T left-- times 1 133 00:06:19,820 --> 00:06:30,130 over T delta T plus 1 over V delta V is equal to-- well, 134 00:06:30,130 --> 00:06:34,650 zero divided by anything is just equal to 0. 135 00:06:34,650 --> 00:06:36,910 Now we're going to integrate over a bunch of really small 136 00:06:36,910 --> 00:06:40,470 delta T's and delta V's. 137 00:06:40,470 --> 00:06:43,250 So let me just change those to our calculus terminology. 138 00:06:43,250 --> 00:06:46,740 We're going to do an infinite sum over infinitesimally small 139 00:06:46,740 --> 00:06:49,410 changes in delta T and delta V. 140 00:06:49,410 --> 00:07:01,030 So I'll rewrite this as 3/2 1 over T dt plus 1 over V dv is 141 00:07:01,030 --> 00:07:01,750 equal to 0. 142 00:07:01,750 --> 00:07:03,860 Remember, this just means a very, very 143 00:07:03,860 --> 00:07:06,690 small change in volume. 144 00:07:06,690 --> 00:07:09,840 This is a very, very, very, small change, an 145 00:07:09,840 --> 00:07:12,580 infinitesimally small change, in temperature. 146 00:07:12,580 --> 00:07:15,540 And now I want to do the total change in temperature. 147 00:07:15,540 --> 00:07:17,240 I want to integrate over the total change in temperature 148 00:07:17,240 --> 00:07:18,960 and the total change in volume. 149 00:07:18,960 --> 00:07:20,880 So let's do that. 150 00:07:20,880 --> 00:07:26,620 So I want to go from always temperature start to 151 00:07:26,620 --> 00:07:27,870 temperature finish. 152 00:07:27,870 --> 00:07:30,010 153 00:07:30,010 --> 00:07:33,060 And this will be going from our volume 154 00:07:33,060 --> 00:07:39,350 start to volume finish. 155 00:07:39,350 --> 00:07:40,630 Fair enough. 156 00:07:40,630 --> 00:07:42,250 Let's do these integrals. 157 00:07:42,250 --> 00:07:45,790 This tends to show up a lot in thermodynamics, these 158 00:07:45,790 --> 00:07:46,620 antiderivatives. 159 00:07:46,620 --> 00:07:49,800 The antiderivative of 1 over T is natural log of T. 160 00:07:49,800 --> 00:07:57,190 So this is equal to 3/2 times the natural log of T. 161 00:07:57,190 --> 00:08:00,870 We're going to evaluate it at the final temperature and then 162 00:08:00,870 --> 00:08:06,370 the starting temperature, plus the natural log-- the 163 00:08:06,370 --> 00:08:09,340 antiderivative of 1 over V is just the natural log of V-- 164 00:08:09,340 --> 00:08:17,120 plus the natural log of V, evaluated from our final 165 00:08:17,120 --> 00:08:18,310 velocity, and we're going to subtract out 166 00:08:18,310 --> 00:08:19,300 the starting velocity. 167 00:08:19,300 --> 00:08:21,400 This is just the calculus here. 168 00:08:21,400 --> 00:08:23,540 And this is going to be equal to 0. 169 00:08:23,540 --> 00:08:23,790 Right? 170 00:08:23,790 --> 00:08:27,610 I mean, we could integrate both sides-- well, if every 171 00:08:27,610 --> 00:08:31,010 infinitesimal change is equal to the sum is equal to 0, the 172 00:08:31,010 --> 00:08:33,039 sum of all of the infinitesimal changes are also 173 00:08:33,039 --> 00:08:33,840 going to be equal to 0. 174 00:08:33,840 --> 00:08:37,460 So this is still equal to 0. 175 00:08:37,460 --> 00:08:38,640 See what we can do here. 176 00:08:38,640 --> 00:08:45,670 So we could rewrite this green part as-- so it's 3/2 times 177 00:08:45,670 --> 00:08:50,170 the natural log of TF minus the natural log of TS, which 178 00:08:50,170 --> 00:08:53,320 is just, using our log properties, the natural log of 179 00:08:53,320 --> 00:08:56,900 TF over the natural log of TS. 180 00:08:56,900 --> 00:08:57,610 Right? 181 00:08:57,610 --> 00:09:00,030 When you evaluate, you get natural log of TF minus the 182 00:09:00,030 --> 00:09:01,180 natural log of TS. 183 00:09:01,180 --> 00:09:03,160 That's the same thing as this. 184 00:09:03,160 --> 00:09:09,820 Plus, for the same reason, the natural log of VF over the 185 00:09:09,820 --> 00:09:12,760 natural log of VS. When you evaluate this, it's the 186 00:09:12,760 --> 00:09:15,890 natural log of VF minus the natural log of VS, which can 187 00:09:15,890 --> 00:09:18,380 be simplified this to this, just from our logarithmic 188 00:09:18,380 --> 00:09:19,560 properties. 189 00:09:19,560 --> 00:09:21,220 So this equals 0. 190 00:09:21,220 --> 00:09:23,940 And now we can-- this coefficient out front, we can 191 00:09:23,940 --> 00:09:25,320 use our logarithmic properties. 192 00:09:25,320 --> 00:09:27,990 Instead of putting a 3/2 natural log of this, we can 193 00:09:27,990 --> 00:09:34,150 rewrite this as the natural log of TF over TS to the 3/2. 194 00:09:34,150 --> 00:09:36,110 Now we can keep doing our logarithm properties. 195 00:09:36,110 --> 00:09:38,350 You take the log of something plus the log of something. 196 00:09:38,350 --> 00:09:40,530 That's equal to the log of their product. 197 00:09:40,530 --> 00:09:45,520 So this is equal to-- I'll switch colors-- The natural 198 00:09:45,520 --> 00:09:56,870 log of TF over TS to the 3/2 power, times the natural log 199 00:09:56,870 --> 00:10:03,090 of VF over VS. And this is a fatiguing proof. 200 00:10:03,090 --> 00:10:03,830 All right. 201 00:10:03,830 --> 00:10:07,730 And all that is going to be equal to 0. 202 00:10:07,730 --> 00:10:09,010 Now what can we say? 203 00:10:09,010 --> 00:10:12,040 Well, we're saying that e to the 0 power-- the natural log 204 00:10:12,040 --> 00:10:16,470 is log base e-- e to the 0 power is equal to this thing. 205 00:10:16,470 --> 00:10:18,410 So this thing must be 1. 206 00:10:18,410 --> 00:10:20,830 E to the 0 power is 1. 207 00:10:20,830 --> 00:10:28,020 So we can say-- we're almost there-- that TF, our final 208 00:10:28,020 --> 00:10:33,900 temperature over our starting temperature to the 3/2 power, 209 00:10:33,900 --> 00:10:38,860 times our final volume over our starting 210 00:10:38,860 --> 00:10:43,400 volume is equal to 1. 211 00:10:43,400 --> 00:10:47,280 Now let's take this result that we worked reasonably hard 212 00:10:47,280 --> 00:10:48,530 to produce. 213 00:10:48,530 --> 00:10:50,430 214 00:10:50,430 --> 00:10:52,940 Remember all of this, we just said, we're dealing with an 215 00:10:52,940 --> 00:10:56,450 adiabatic process, and we started from the principle of 216 00:10:56,450 --> 00:10:58,890 just what the definition of internal energy is. 217 00:10:58,890 --> 00:11:03,150 And then we substitute it with our PV equals nRT formulas. 218 00:11:03,150 --> 00:11:06,090 Although this was kind of PV-- this is internal energy at any 219 00:11:06,090 --> 00:11:09,290 point is equal to 3/2 times PV. 220 00:11:09,290 --> 00:11:11,920 And then we integrated over all the changes, and we said, 221 00:11:11,920 --> 00:11:13,240 look, this is adiabatic. 222 00:11:13,240 --> 00:11:17,410 So the total change-- the sum of all of our change in 223 00:11:17,410 --> 00:11:19,780 internal energy and work done by the system has to be 0, 224 00:11:19,780 --> 00:11:22,560 then we use the property of log to get to this result. 225 00:11:22,560 --> 00:11:24,850 Now let's do these for both of these adiabatic 226 00:11:24,850 --> 00:11:26,890 processes over here. 227 00:11:26,890 --> 00:11:30,110 So the first one we could do is this one where we go from 228 00:11:30,110 --> 00:11:35,690 volume B at T1 to volume C at T2. 229 00:11:35,690 --> 00:11:40,370 Watch the Carnot cycle video, if you forgot that. 230 00:11:40,370 --> 00:11:42,860 This was the VB All of these things up here were at 231 00:11:42,860 --> 00:11:43,690 temperature 1. 232 00:11:43,690 --> 00:11:46,206 All of the things down here were at temperature 2 So we're 233 00:11:46,206 --> 00:11:49,480 at temperature 1 up here, and temperature 2 down 234 00:11:49,480 --> 00:11:50,940 here, volume C. 235 00:11:50,940 --> 00:11:51,840 So let's look at that. 236 00:11:51,840 --> 00:11:57,850 So on that right part, that right process, our final 237 00:11:57,850 --> 00:11:59,880 temperature was temperature 2. 238 00:11:59,880 --> 00:12:00,900 So let me write it down. 239 00:12:00,900 --> 00:12:02,710 Temperature 2. 240 00:12:02,710 --> 00:12:04,875 Our initial temperature was temperature 1, where we 241 00:12:04,875 --> 00:12:09,110 started off at point B to the 3/2. 242 00:12:09,110 --> 00:12:11,660 Times-- what was our final volume? 243 00:12:11,660 --> 00:12:15,040 Our final volume was our volume at C divided by the 244 00:12:15,040 --> 00:12:16,680 volume at B. 245 00:12:16,680 --> 00:12:19,460 And that's going to be equal to 1. 246 00:12:19,460 --> 00:12:20,600 Neat. 247 00:12:20,600 --> 00:12:23,260 That's the result we got from this adiabatic process. 248 00:12:23,260 --> 00:12:25,370 We got that formula saying, this is adiabatic, we did a 249 00:12:25,370 --> 00:12:27,380 bunch of math, and then we just substitute for our 250 00:12:27,380 --> 00:12:30,990 initial and final volumes and temperatures. 251 00:12:30,990 --> 00:12:35,310 Let's do it the same way, but let's go from D to A. 252 00:12:35,310 --> 00:12:42,760 So when you go from D to A, what's your final temperature? 253 00:12:42,760 --> 00:12:44,330 Don't want to get you dizzy going up. 254 00:12:44,330 --> 00:12:46,320 Well your final temperature, we're going from D to A. 255 00:12:46,320 --> 00:12:48,150 So our final temperature is T1. 256 00:12:48,150 --> 00:12:51,940 and our final volume is the volume at A. 257 00:12:51,940 --> 00:12:53,480 Go back down. 258 00:12:53,480 --> 00:12:56,580 So our final temperature is T1. 259 00:12:56,580 --> 00:12:58,690 Our initial temperature is T2. 260 00:12:58,690 --> 00:13:06,430 We're going from D to A to the 3/2 power is times-- let me 261 00:13:06,430 --> 00:13:07,520 write our form formula there. 262 00:13:07,520 --> 00:13:10,250 Our final volume is the volume at A. 263 00:13:10,250 --> 00:13:12,630 That's where we moved to. 264 00:13:12,630 --> 00:13:14,380 And we moved from our volume at D. 265 00:13:14,380 --> 00:13:16,580 And this is going to be equal to 1. 266 00:13:16,580 --> 00:13:16,990 OK. 267 00:13:16,990 --> 00:13:20,110 We're almost there, if your eyes are 268 00:13:20,110 --> 00:13:21,310 beginning to glaze over. 269 00:13:21,310 --> 00:13:22,280 But this is interesting. 270 00:13:22,280 --> 00:13:25,490 And if anything, it's a little bit of fun mathematics to wake 271 00:13:25,490 --> 00:13:26,760 you up in the morning. 272 00:13:26,760 --> 00:13:27,910 So let's see. 273 00:13:27,910 --> 00:13:29,890 We can almost relate these two things. 274 00:13:29,890 --> 00:13:31,880 We could set them equal to each other, but it's not quite 275 00:13:31,880 --> 00:13:33,690 satisfying yet. 276 00:13:33,690 --> 00:13:38,670 Let's take the reciprocal of both sides of this equation 277 00:13:38,670 --> 00:13:39,710 right here. 278 00:13:39,710 --> 00:13:44,890 So obviously if we take the reciprocal of this-- and we 279 00:13:44,890 --> 00:13:47,990 could just say, this is T2 over T1 to the minus 3/2 280 00:13:47,990 --> 00:13:53,670 power, which is the same thing as T1 over 281 00:13:53,670 --> 00:13:56,580 T2, to the 3/2 power. 282 00:13:56,580 --> 00:13:56,780 Right? 283 00:13:56,780 --> 00:13:58,000 That's just the reciprocal. 284 00:13:58,000 --> 00:14:00,540 And we're taking both sides to the negative 1 power, so we're 285 00:14:00,540 --> 00:14:01,990 going to have to take this to the negative 1 power. 286 00:14:01,990 --> 00:14:04,635 VB over VC. 287 00:14:04,635 --> 00:14:06,390 And when you take the reciprocal of 1, that just 288 00:14:06,390 --> 00:14:09,530 equals 1, That still equals 1. 289 00:14:09,530 --> 00:14:12,230 Which this also equals, so we could say, that equals this 290 00:14:12,230 --> 00:14:13,560 thing over here. 291 00:14:13,560 --> 00:14:19,550 So that is equal to T1 over T2, to the 3/2 power, 292 00:14:19,550 --> 00:14:24,180 times VA over VD. 293 00:14:24,180 --> 00:14:25,600 Now, these things are equal to each other. 294 00:14:25,600 --> 00:14:26,500 We can get rid of the 1. 295 00:14:26,500 --> 00:14:30,120 These two-- actually, let me just erase some of this. 296 00:14:30,120 --> 00:14:31,580 I don't want to make it say not equal to. 297 00:14:31,580 --> 00:14:32,540 They're equal to each other. 298 00:14:32,540 --> 00:14:34,980 They both equal 1. 299 00:14:34,980 --> 00:14:36,960 So they both equal each other. 300 00:14:36,960 --> 00:14:40,070 301 00:14:40,070 --> 00:14:41,960 This thing and this thing are the same thing. 302 00:14:41,960 --> 00:14:44,700 T1 over T2 to the 3/2, T1 over T2 to the 3/2. 303 00:14:44,700 --> 00:14:47,370 So let's just divide both sides by that. 304 00:14:47,370 --> 00:14:48,750 Those cancel out. 305 00:14:48,750 --> 00:14:50,190 And what are we left with? 306 00:14:50,190 --> 00:14:52,380 I think you can see the finish line. 307 00:14:52,380 --> 00:14:54,140 The finish line is near. 308 00:14:54,140 --> 00:15:02,730 We have VB over VC is equal to VA over VD. 309 00:15:02,730 --> 00:15:04,740 Now that's not quite the result we wanted, but it takes 310 00:15:04,740 --> 00:15:07,300 a little bit of simple arithmetic to get there. 311 00:15:07,300 --> 00:15:09,240 Let's just cross-multiply. 312 00:15:09,240 --> 00:15:18,160 And you get VB times VD is equal to VC times VA. 313 00:15:18,160 --> 00:15:30,710 Now if we divide both sides by VBVC-- actually, let's do it 314 00:15:30,710 --> 00:15:31,070 the other way. 315 00:15:31,070 --> 00:15:32,350 Let's divide both sides by VDVA. 316 00:15:32,350 --> 00:15:40,010 317 00:15:40,010 --> 00:15:41,340 What do we get? 318 00:15:41,340 --> 00:15:44,780 These cancel out, and these cancel out. 319 00:15:44,780 --> 00:15:53,870 And we are left with VB over VA is equal to VC over VD. 320 00:15:53,870 --> 00:15:56,800 All that work for a nice and simple result, but that's 321 00:15:56,800 --> 00:15:59,620 better than doing a lot of work for a hairy 322 00:15:59,620 --> 00:16:01,140 and monstrous result. 323 00:16:01,140 --> 00:16:03,440 So that's what we set out to prove. 324 00:16:03,440 --> 00:16:06,960 That VB over VA is equal to VC over VD. 325 00:16:06,960 --> 00:16:09,190 And we got it all from the notion that we're dealing with 326 00:16:09,190 --> 00:16:13,220 adiabatic process, that our change in heat is 0. 327 00:16:13,220 --> 00:16:16,080 and we just went to our formula, or our definition of 328 00:16:16,080 --> 00:16:18,260 our change in internal energy, the first law of 329 00:16:18,260 --> 00:16:21,130 thermodynamics, that if we have any change in internal 330 00:16:21,130 --> 00:16:23,450 energy, it must be equal to the amount of 331 00:16:23,450 --> 00:16:25,470 work done by the system. 332 00:16:25,470 --> 00:16:27,880 Or at least a negative of the work done by the system. 333 00:16:27,880 --> 00:16:29,720 When you add them up, you get to 0. 334 00:16:29,720 --> 00:16:33,120 Then we use that result from a few videos ago, where we said 335 00:16:33,120 --> 00:16:36,660 the internal energy at any point is 3/2 times nRT. 336 00:16:36,660 --> 00:16:39,490 So the change in internal energy is that times delta T, 337 00:16:39,490 --> 00:16:41,210 because that's the only thing that can change. 338 00:16:41,210 --> 00:16:42,890 We used PV equal nRT. 339 00:16:42,890 --> 00:16:45,430 And then we just integrated along all of the little 340 00:16:45,430 --> 00:16:47,290 changes in temperature and volume, as we 341 00:16:47,290 --> 00:16:49,030 moved along this line. 342 00:16:49,030 --> 00:16:51,700 As we moved along the line, we took the integral. 343 00:16:51,700 --> 00:16:54,110 We said that had to be equal to 0. 344 00:16:54,110 --> 00:16:56,110 And we ended up with this formula over here, and then we 345 00:16:56,110 --> 00:16:58,010 just applied it to our two adiabatic processes. 346 00:16:58,010 --> 00:17:01,580 And we went from B to C, and we went from D to A. 347 00:17:01,580 --> 00:17:02,695 And we got these results. 348 00:17:02,695 --> 00:17:05,848 And we got to our finish line. 349 00:17:05,848 --> 00:00:00,000 See you in the next video.